The Hilbert cube Q is the subset of l∞ given by {a : N→R││a(i)│≤2-i}.

Show that Q is a compact subspace of l∞ (with the norm ∥a∥=supn∈N│a(n)│).

can anyone please help me to do this question ?? ,,

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- May 3rd 2009, 03:25 AMjin_nzzangcompactness
The Hilbert cube Q is the subset of l∞ given by {a : N→R││a(i)│≤2-i}.

Show that Q is a compact subspace of l∞ (with the norm ∥a∥=supn∈N│a(n)│).

can anyone please help me to do this question ?? ,,

- May 4th 2009, 09:40 AMOpalg
Let $\displaystyle \mathcal B$ denote the unit ball of $\displaystyle l^\infty$. Then $\displaystyle \mathcal B$ consists of all sequences $\displaystyle (x_n)$ of real numbers such that $\displaystyle x_n\in[-1,1]$ for all n. Give $\displaystyle \mathcal B$ the topology that it has as the product of an infinite number of copies of [–1,1]. In this topology, a sequence $\displaystyle ((x^{(k)}))$ converges to $\displaystyle ((x^{(0)}))$ if it converges coordinatewise. That is to say, if $\displaystyle x^{(k)}_n\to x^{(0)}_n$ for each coordinate n.

The map $\displaystyle f:(x_n)\mapsto(2^{-n}x_n)$ takes $\displaystyle \mathcal B$ onto Q. The idea of the proof is to show that f is continuous from $\displaystyle \mathcal B$ (with the above product topology) to Q (with the topology from the $\displaystyle l^\infty$ norm). By Tychonoff's theorem, any product of compact spaces is compact. So $\displaystyle \mathcal B$ is compact, and hence Q, as a continuous image of a compact space, is also compact.

To see that f is continuous, suppose that $\displaystyle x^{(k)}\to x^{(0)}$ in $\displaystyle \mathcal B$, and let $\displaystyle \varepsilon>0$. Choose N so that $\displaystyle 2^{-N}<\varepsilon/2$. Then $\displaystyle |(f(x^{(k)}) - f(x^{(0)}))_n| = 2^{-n}|x^{(k)}_n - x^{(0)}_n| < \varepsilon$ for all n>N. But (by the definition of the product topology in $\displaystyle \mathcal B$) there exists K such that $\displaystyle |x^{(k)}_n - x^{(0)}_n| < \varepsilon$ for all k>K and all n with 1≤n≤N. Therefore $\displaystyle \|f(x^{(k)}) - f(x^{(0)})\|_\infty \leqslant \varepsilon$, so that $\displaystyle f(x^{(k)})\to f(x^{(0)})$ in Q.

[That is essentially the argument given here to show that the Hilbert cube is compact for the topology that comes from the $\displaystyle l^2$-norm, except that the Wikipedia page completely ducks the central part of the proof, which is to show that the map f is continuous.]